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3 years ago

Int[] array1 = {1, 3, 5, 7}

Computers and Technology

1 answer:

Helga [31]3 years ago

4 0

Answer: 0,1,2,3

Explanation:

The array contains 4 elements that is 1,3,5,7.So the size of the array is 4.The indexing in an array starts from 0 means the element at 0 index is the first element in the array So 1 is at the index 0 and the index will increase by 1 as we move right of the array.So 3 is at the index 1 ,5 is at the index 2 and 7 is at the index 3.So the indices are 0,1,2,3.

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3 years ago

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Answer:

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3 years ago

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4 years ago

Suppose you have two arrays of ints, arr1 and arr2, each containing ints that are sorted in ascending order. Write a static meth

telo118 [61]

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public static int[] merge(int[] arr1, int[] arr2) {
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int[] result = new int[arr1.length+arr2.length]
while(i < arr1.length && j < arr2.length) {
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if(arr1[i] < arr2[j]) {
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}
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for(int index = isArr1 ? i : j; index < isArr1 ? arr1.length : arr2.length; index++) {
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So this implementation is kind of confusing, but it's the first way I thought to do it so I ran with it. There is probably an easier way, but that's the beauty of programming.

A quick explanation:

We first loop through the arrays comparing the first elements of each array, adding whichever is the smallest to the result array. Each time we do so, we increment the index value (i or j) for the array that had the smaller number. Now the next time we are comparing the NEXT element in that array to the PREVIOUS element of the other array. We do this until we reach the end of either arr1 or arr2 so that we don't get an out of bounds exception.

The second step in our method is to tack on the remaining integers to the resulting array. We need to do this because when we reach the end of one array, there will still be at least one more integer in the other array. The boolean isArr1 is telling us whether arr1 is the array with leftovers. If so, we loop through the remaining indices of arr1 and add them to the result. Otherwise, we do the same for arr2. All of this is done using ternary operations to determine which array to use, but if we wanted to we could split the code into two for loops using an if statement.

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4 years ago

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alina1380 [7]

Answer:

it can balance computing resources by time

Explanation:

this this is as it involves multiprogramming

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4 years ago